Problem: Simplify and expand the following expression: $ \dfrac{3k - 3}{5k - 1}-\dfrac{2k}{2k + 10} $
Solution: In order to subtract expressions, they must have a common denominator. Get both fractions over a common denominator of $(5k - 1)(2k + 10)$ Multiply the first term by $\dfrac{2k + 10}{2k + 10}$ $ \begin{align*} \dfrac{3k - 3}{5k - 1} \times \dfrac{2k + 10}{2k + 10} & = \dfrac{(3k - 3)(2k + 10)}{(5k - 1)(2k + 10)} \\ & = \dfrac{6k^2 + 24k - 30}{(5k - 1)(2k + 10)}\end{align*} $ Multiply the second term by $\dfrac{5k - 1}{5k - 1}$ $ \begin{align*} \dfrac{2k}{2k + 10} \times \dfrac{5k - 1}{5k - 1} & = \dfrac{(2k)(5k - 1)}{(2k + 10)(5k - 1)} \\ & = \dfrac{10k^2 - 2k}{(2k + 10)(5k - 1)}\end{align*} $ Now we have: $ = \dfrac{6k^2 + 24k - 30}{(5k - 1)(2k + 10)} - \dfrac{10k^2 - 2k}{(2k + 10)(5k - 1)} $ Now both terms have a common denominator we can subtract the numerators: $ = \dfrac{6k^2 + 24k - 30 - (10k^2 - 2k)}{(5k - 1)(2k + 10)} $ $ = \dfrac{6k^2 + 24k - 30 - 10k^2 + 2k}{(5k - 1)(2k + 10)} $ $ = \dfrac{-4k^2 + 26k - 30}{(5k - 1)(2k + 10)}$ Expand the denominator: $ = \dfrac{-4k^2 + 26k - 30}{10k^2 + 48k - 10}$ Simplify: $ = \dfrac{-2k^2 + 13k - 15}{5k^2 + 24k - 5}$